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A load of 10 KN is supported from a pul...

A load of `10 KN` is supported from a pulley , which in turn is supported by a rope of cross-sectional area `10^(3) mm^(2)` and modulus of elasticity `10^(3) Nmm^(-2)` as shown in the figure. Neglecting friction at the pulley , then downward deflection of the load (in mm) is

Text Solution

Verified by Experts

Let `T` be the tension in the rope. Then
`2T=10kNimpliesT=5N`

Hence longitudinal stress in the rope is
`sigma=T/A=(5kN)/(10^(3)mm^(2))=5N//mm^(2)`
Extension in the rope `=("Stress")/YxxL`
`=-(5N//mm^(2))/(10^(3)N//mm)xx1500mm=7.5mm`
Therefore, deflection of the load `delta=7.5/2=3.75mm`
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